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JEE Exam  >  JEE Questions  >  A proton, a neutron, an electron and an &alph... Start Learning for Free
A proton, a neutron, an electron and an α-particle have same energy. If λp, λn, λe and λα are the de-Broglie's wavelengths of proton, neutron, electron and α particle, respectively then choose the correct relation from the following:
  • a)
    λp = λn > λe > λα
  • b)
    λα < λn < λp < λe
  • c)
    λe < λp = λn > λα
  • d)
    λe = λp = λn = λα
Correct answer is option 'B'. Can you explain this answer?
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Most Upvoted Answer
A proton, a neutron, an electron and an α-particle have same ene...
Alpha particle walk into a bar.

The proton orders a pint of beer and starts chatting with the bartender about the latest scientific discoveries. The neutron orders a glass of water and joins the conversation, adding some interesting facts about nuclear physics.

The electron, being a bit of a troublemaker, orders a shot of tequila and starts dancing on the bar counter, creating a vibrant light show. The bartender tries to calm the electron down, but it's too late. The electron's wild dance moves knock over some glasses and cause a small electrical surge.

Just as the bartender is about to kick the electron out, the alpha particle walks in. The alpha particle is composed of two protons and two neutrons tightly bound together. It takes one look at the chaos caused by the electron and immediately goes into action.

The alpha particle gently approaches the electron and starts pulling it away from the bar, using its strong nuclear force to keep the electron in check. The electron, realizing it's outmatched, reluctantly follows the alpha particle's lead.

The bartender breathes a sigh of relief as the alpha particle successfully escorts the electron out of the bar, restoring peace and order. The proton and neutron thank the alpha particle for its help and continue their scientific discussion with the bartender, now enjoying a calm and quiet evening at the bar.
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Community Answer
A proton, a neutron, an electron and an α-particle have same ene...
de Broglie wavelength λ = h/p

Where K: kinetic energy
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The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. If the uncertainty in velocity position is same, then the uncertainty in momentum will be

Can you explain the answer of this question below:The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q.The correct order of wavelength of Hydrogen (1H1), Deuterium (1H2) and Tritium (1H3) moving with same kinetic energy is :A:H D TB:H = D = TC:H D TD:H D TThe answer is a.

The French physicist Louis de-Broglie in 1924 postulated that matter, like radiation, should exhibit a dual behaviour. He proposed the following relationship between the wavelength of a material particle, its linear momentum p and planck constant h.The de Broglie relation implies that the wavelength of a particle should decreases as its velocity increases. It also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de Broglie waves.These waves differ from the electromagnetic waves as they,(i) have lower velocities(ii) have no electrical and magnetic fields and(iii) are not emitted by the particle under consideration.The experimental confirmation of the deBroglie relation was obtained when Davisson and Germer, in 1927, observed that a beam of electrons is diffracted by a nickel crystal. As diffraction is a characteristic property of waves, hence the beam of electron behaves as a wave, as proposed by deBroglie.Werner Heisenberg considered the limits of how precisely we can measure properties of an electron or other microscopic particle like electron. He determined that there is a fundamental limit of how closely we can measure both position and momentum. The more accurately we measure the momentum of a particle, the less accurately we can determine its position. The converse is also ture. This is summed up in what we now call the Heisenberg uncertainty principle : It is impossible to determine simultaneously and precisely both the momentum and position of a particle. The product of undertainty in the position, x and the uncertainity in the momentum (mv) must be greater than or equal to h/4. i.e.Q. The transition, so that the de - Broglie wavelength of electron becomes 3 times of its initial value in He+ ion will be

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A proton, a neutron, an electron and an α-particle have same energy. If λp,λn,λeandλα are the de-Broglies wavelengths of proton, neutron, electron and αparticle, respectively then choose the correct relation from the following:a)λp =λn >λe >λαb)λα< λn< λp< λec)λe <λp =λn >λαd)λe =λp =λn =λαCorrect answer is option 'B'. Can you explain this answer?
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A proton, a neutron, an electron and an α-particle have same energy. If λp,λn,λeandλα are the de-Broglies wavelengths of proton, neutron, electron and αparticle, respectively then choose the correct relation from the following:a)λp =λn >λe >λαb)λα< λn< λp< λec)λe <λp =λn >λαd)λe =λp =λn =λαCorrect answer is option 'B'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A proton, a neutron, an electron and an α-particle have same energy. If λp,λn,λeandλα are the de-Broglies wavelengths of proton, neutron, electron and αparticle, respectively then choose the correct relation from the following:a)λp =λn >λe >λαb)λα< λn< λp< λec)λe <λp =λn >λαd)λe =λp =λn =λαCorrect answer is option 'B'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A proton, a neutron, an electron and an α-particle have same energy. If λp,λn,λeandλα are the de-Broglies wavelengths of proton, neutron, electron and αparticle, respectively then choose the correct relation from the following:a)λp =λn >λe >λαb)λα< λn< λp< λec)λe <λp =λn >λαd)λe =λp =λn =λαCorrect answer is option 'B'. Can you explain this answer?.
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